document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1151193>Question 1151193</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I drove to my parents' house at $40$ miles per hour. If I had driven 30 miles an hour instead, then I would have gotten there 20 minutes later. How many miles did I drive? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #772856 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13203)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=772856\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "The solution method shown by tutor @josgarithmetic is a perfectly good standard algebraic method.<br><BR>\n" );
document.write( "If you are one of many who dislike working with fractions, here is a very different, somewhat informal method.<br><BR>\n" );
document.write( "The ratio of the two speeds is 40:30 = 4:3; since the distances are the same, the ratio of the times is 3:4.<br><BR>\n" );
document.write( "Given that ratio of times, let the two times be 3x and 4x.  The difference, 4x-3x=x, is 20 minutes.<br><BR>\n" );
document.write( "That means the amount of time at 40mph was 3x = 3(20 minutes) = 60 minutes = 1 hour.  And 40mph for 1 hour makes the distance 40 miles.<br><BR>\n" );
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